A person's savings for 3 years are in an arithmetic progression that in the 3 years he has saved cs 24,000 and in the first year he saves more than he saved in the second year

Let's denote the amount saved in the first year as a , the common difference as d , and the total amount saved in 3 years as 24,000 . Since the savings are in an arithmetic progression, we have the following equations:

1) Amount saved in the 1st year: a
2) Amount saved in the 2nd year: a + d
3) Amount saved in the 3rd year: a + 2d

From the given information, we have the total amount saved in 3 years:
a + (a + d) + (a + 2d) = 24,000

Simplify the equation:
3a + 3d = 24,000

Divide by 3:
a + d = 8,000

Since the amount saved in the 1st year is more than the 2nd year, we have:
a > a + d

Simplify this inequality:
a > 8,000

Therefore, the person's savings for the 3 years are a + (a+d) + (a+2d) = 2a + 3d = a + a + 8,000 = 2a + 8,000 and the amount saved in the first year is greater than 8,000 .

\textbf{Answer:} The person's savings for 3 years are 2a + 8,000 , where a > 8,000 .